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$xhtml = array(
	'<{title}>' => 'Trigonometry',
	'<{subtitle}>' => 'Written in <span title="College Algebra">MATH 1201</span> by <a href="https://y.st./">Alex Yst</a>, finalised on 2018-03-21',
	'<{copyright year}>' => '2018',
	'takedown' => '2017-11-01',
	'<{body}>' => <<<END
<section id="problem0">
	<h2>Problem 0</h2>
	<p>
		(As a side note, the answer to this problem, by definition, is (cos(330°), sin(330°)).
		All you have to do is type that into a calculator.)
	</p>
	<p>
		There are 360° in a circle, so if we subtract 330°, we have 30°, the difference between a full circle and the angle we&apos;re working with.
		The textbook tells us that a 30° angle is π÷6 radians.
		A full circle is 2π radians, so we can subtract π÷6 radians from 2π radians to arrive at 11π÷2 radians.
		In doing so, we&apos;ve just converted our initial angle&apos;s measure into radians: 330° = 11π÷6 radians.
		11π÷6 radians is one of the special values on the unit circle that the textbook told us to memorise.
		The point we memorised as being the one this angle intersects on the unit circle is (√3÷2, -½).
	</p>
</section>
<section id="problem1">
	<h2>Problem 1</h2>
	<p>
		The first thing we want to do is get the cosine term onto one side of the equation by itself.
		We can do this by subtracting three from both sides:
	</p>
	<p>
		3 + cos(2x) = 7÷2<br/>
		cos(2x) = ½
	</p>
	<p>
		Again, referring to the unit circle we memorised, only two values within the specified range, [0, 2π], have a cosine of ½: π÷3 and 5π÷3.
		We need to remember that we&apos;re still looking at 2x though, not x, so we need to double that range for the time being.
		We add 2π to both values to get another two values to use alongside them.
		To put it more clearly:
	</p>
	<p>
		cos(π÷3) = ½, cos(5π÷3) = ½, cos(7π÷3) = ½, and cos(11π÷3) = ½
	</p>
	<p>
		We still have the fact that we have cos(2x) though, not cos(x), as stated above.
		We&apos;ve got to divide our answers by two.
		We can easily do this by just doubling the denominator, and all four solutions are within the specified range of [0, 2π].
		Doing that, we arrive at our final solution: <strong>x = {π÷6, 5π÷6, 7π÷6, 11π÷6}</strong>.
	</p>
</section>
<section id="problem2">
	<h2>Problem 2</h2>
	<p>
		To prove this identity:
	</p>
	<p>
		cos(2x) = 1 - 2sin<sup>2</sup>(x)
	</p>
	<p>
		We&apos;ll use another identity provided by the book:
	</p>
	<p>
		cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
	</p>
	<p>
		In the identity we are to prove, 2x simply means that a ans b both have the same value; namely, x.
		So:
	</p>
	<p>
		cos(2x) = cos(x)cos(x) - sin(x)sin(x)
	</p>
	<p>
		Right away, we can see that the sines and cosigns here are being multiplied by themselves.
		They&apos;re being squared.
		So we rewrite as:
	</p>
	<p>
		cos<sup>2</sup>(x) - sin<sup>2</sup>(x)
	</p>
	<p>
		We also know that the very definition of sine and cosine are such that cos<sup>2</sup>(x) plus sin<sup>2</sup>(x) is always equal to one.
		In other words:
	</p>
	<p>
		1 - cos<sup>2</sup>(x) - sin<sup>2</sup>(x) = 0
	</p>
	<p>
		We can add zero to any value without altering said value, so:
	</p>
	<p>
		cos<sup>2</sup>(x) - sin<sup>2</sup>(x) = cos<sup>2</sup>(x) - sin<sup>2</sup>(x) + 1 - cos<sup>2</sup>(x) - sin<sup>2</sup>(x)
	</p>
	<p>
		We can simplify this by combining like terms.
		cos<sup>2</sup>(x) - cos<sup>2</sup>(x) = 0 and -sin<sup>2</sup>(x) - sin<sup>2</sup>(x) = -2sin<sup>2</sup>(x), so:
	</p>
	<p>
		cos<sup>2</sup>(x) - sin<sup>2</sup>(x) + 1 - cos<sup>2</sup>(x) - sin<sup>2</sup>(x) = 1 - 2sin<sup>2</sup>(x)
	</p>
	<p>
		We started with cos(2x) and worked our way to 1 - 2sin<sup>2</sup>(x), proving that the two are equivalent.
	</p>
</section>
END
);
